Solved Problem on Dynamics
Português
English
In the system of the figure, the body B slides on a horizontal surface without friction, it is connected through
by ropes and pulleys, lightweight and frictionless, with two bodies A and C that move vertically. The
masses of A, B, and C are, respectively, 5 kg, 2 kg and 3 kg. Find the acceleration of the system
and the tension in the cord. Assume
g = 10 m/s2.
Problem data
- mass of body A: mA = 5 kg;
- mass of body B: mB = 2 kg;
- mass of body C: mC = 3 kg;
- free-fall acceleration: g = 10 m/s2.
Problem diagram
We choose the acceleration in the direction in which the body A moves downward. Drawing a free-body diagram for each block and using Newton's Second Law
We choose the acceleration in the direction in which the body A moves downward. Drawing a free-body diagram for each block and using Newton's Second Law
\[ \bbox[#99CCFF,10px]
{\vec{F}=m\vec{a}}
\]
Solution
Body A
Body A
- \( \vec W_{\text{A}} \): weight of body A;
- \( \vec T_{\text{AB}} \): tension in the cord between blocks A and B.
figure 2
\[
P_{\text{A}}-T_{\text{AB}}=m_{\text{A}}a \tag{I}
\]
Body B
Vertical direction
Vertical direction
- \( \vec W_{\text{B}} \): weight of body B;
- \( \vec N_{\text{B}} \): normal force of the surface on the body.
- \( \vec T_{\text{AB}} \): tension in the rope between blocks A and B;
- \( \vec T_{\text{BC}} \): tension in the rope between blocks B and C.
figure 3
In the vertical direction, the weight and the normal force cancel out, there is no vertical motion.
In the horizontal direction by using Newton's Second Law we have
In the horizontal direction by using Newton's Second Law we have
\[
T_{\text{AB}}-T_{\text{BC}}=m_{\text{B}}a \tag{II}
\]
Corpo C
- \( \vec W_{\text{C}} \): weight of body C;
- \( \vec T_{\text{BC}} \): tension in the rope between blocks B and C.
figure 4
\[
T_{\text{BC}}-W_{\text{C}}=m_{\text{C}}a \tag{III}
\]
With equations (I), (II) and (III) we have a system of linear equations with three variables (TAB,
TBC and a), adding the three equations we have
\[
\frac{
\left\{
\begin{array}{rr}
W_{\text{A}}-\cancel{T_{\text{AB}}}&=m_{\text{A}}a\\
\cancel{T_{\text{AB}}}-\cancel{T_{\text{BC}}}&=m_{\text{B}}a\\
\cancel{T_{\text{BC}}}-W_{\text{C}}&=m_{\text{C}}a
\end{array}
\right.}
{W_{\text{A}}-W_{\text{C}}=\left(m_{\text{A}}+m_{\text{B}}+m_{\text{C}}\right)a}
\]
\[
a=\frac{W_{\text{A}}-W_{\text{C}}}{m_{\text{A}}+m_{\text{B}}+m_{\text{C}}} \tag{IV}
\]
The weight of bodies A and C are given by
\[
W_{\text{A}}=m_{\text{A}}g\ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ W_{\text{C}}=m_{\text{C}}g \tag{V}
\]
substituting expression (V) in (IV) and the values given in the problem
\[
a=\frac{m_{\text{A}}g-m_{\text{C}}g}{m_{\text{A}}+m_{\text{B}}+m_{\text{C}}}\\
a=\frac{5 \times 10-3 \times 10}{5+2+3}\\
a=\frac{50-30}{10}\\
a=\frac{20}{10}
\]
\[ \bbox[#FFCCCC,10px]
{a=2\;\text{m/s}^{2}}
\]
Substituting the mass of the body A, the expression for the weight of A given in (V) and the acceleration,
found above, in the first expression of the system, the tension in the cord is
\[
m_{\text{A}}g-T_{\text{AB}}=m_{\text{A}}a\\
5 \times 10-T_{\text{AB}}=5 \times 2\\
50-T_{\text{AB}}=10\\
T_{\text{AB}}=50-10
\]
\[ \bbox[#FFCCCC,10px]
{T_{\text{AB}}=40\ \text{N}}
\]
Substituting the mass of the body C, the expression for the weight of C given in (V) and the acceleration,
found above, in the third expression of the system, the tension in the cord is
\[
T_{\text{BC}}-m_{\text{C}}g=m_{\text{C}}a\\
T_{\text{BC}}-3.10=3.2\\
T_{\text{BC}}-30=6\\
T_{\text{BC}}=6+30
\]
\[ \bbox[#FFCCCC,10px]
{T_{\text{AB}}=36\ \text{N}}
\]
advertisement
advertisement
advertisement
advertisement
advertisement
advertisement
Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .